If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant is

Options

A. `5sqrt2,pi//3`
B. `5sqrt2,pi//4`
C. `-5sqrt2,pi//3`
D. `-5sqrt2,pi//4`

Answer

To solve the problem of finding the magnitude and direction of the resultant of two forces of 5 N each acting along the X and Y axes, we can follow these steps:

### Step 1: Identify the Forces
We have two forces:
- \( F_1 = 5 \, \text{N} \) acting along the X-axis (let's denote this as the i direction).
- \( F_2 = 5 \, \text{N} \) acting along the Y-axis (let's denote this as the j direction).

### Step 2: Represent the Forces as Vectors
We can represent these forces as vectors:
- \( \vec{F_1} = 5 \hat{i} \)
- \( \vec{F_2} = 5 \hat{j} \)

### Step 3: Add the Vectors
To find the resultant vector \( \vec{R} \), we add the two vectors:
\[ \vec{R} = \vec{F_1} + \vec{F_2} = 5 \hat{i} + 5 \hat{j} \]

### Step 4: Calculate the Magnitude of the Resultant
The magnitude of the resultant vector \( \vec{R} \) can be calculated using the Pythagorean theorem:
\[ |\vec{R}| = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, \text{N} \]

### Step 5: Calculate the Direction of the Resultant
To find the direction of the resultant vector, we can use the tangent of the angle \( \theta \) it makes with the X-axis:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{F_2}{F_1} = \frac{5}{5} = 1 \]
Thus, \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \) radians or \( 45^\circ \).

### Final Result
The magnitude of the resultant force is \( 5\sqrt{2} \, \text{N} \) and the direction is \( 45^\circ \) (or \( \frac{\pi}{4} \) radians) from the X-axis.

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